| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | First occurrence probability |
| Difficulty | Standard +0.3 This is a standard tree diagram question requiring systematic application of conditional probability rules across multiple stages. Part (a) involves a single path calculation, part (b) requires identifying one specific path (no rain Sunday/Monday, rain Tuesday), and part (c) needs consideration of four mutually exclusive cases. While it requires careful organization and multiple calculations, the techniques are routine for AS-level statistics with no novel problem-solving insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{no rain}) = 0.6 \times (0.8)^3 = 0.3072,\ \frac{192}{625}\) | B1 | Exact value required |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.6 \times 0.8 \times 0.2\) | M1 | \(a \times b \times c\) where \(a, b = 0.6, 0.8,\ c = 0.2, 0.4, 0.7\). Condone including Wednesday with both 0.3 and 0.7 used |
| \(= 0.096,\ \frac{12}{125}\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(RDDD) = 0.4 \times 0.3 \times 0.8 \times 0.8 = 0.0768,\ \frac{48}{625}\) | B1 | Correct probability for one clearly identified outcome evaluated; accept unsimplified. A correct unsimplified expression is not sufficient |
| \(P(DRDD) = 0.6 \times 0.2 \times 0.3 \times 0.8 = 0.0288,\ \frac{18}{625}\) | M1 | Add 4 probability values, \(0 < p < 1\), for appropriate identified scenarios. Accept unsimplified. Ways of identifying scenarios: stating the days; all unsimplified probability calculations exactly as stated; identifying correct branches on tree diagram and linking with values. No repeated scenarios. No incorrect scenarios |
| \(P(DDRD) = 0.6 \times 0.8 \times 0.2 \times 0.3 = 0.0288,\ \frac{18}{625}\) | ||
| \(P(DDDR) = 0.6 \times 0.8 \times 0.8 \times 0.2 = 0.0768,\ \frac{48}{625}\) | ||
| \(0.2112,\ \frac{132}{625}\) | A1 | Accept 0.211. If 0/3 scored SC B1 for \(0.2112,\ \frac{132}{625}\) |
| Total | 3 |
## Question 2:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{no rain}) = 0.6 \times (0.8)^3 = 0.3072,\ \frac{192}{625}$ | B1 | Exact value required |
| **Total** | **1** | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 \times 0.8 \times 0.2$ | M1 | $a \times b \times c$ where $a, b = 0.6, 0.8,\ c = 0.2, 0.4, 0.7$. Condone including Wednesday with both 0.3 and 0.7 used |
| $= 0.096,\ \frac{12}{125}$ | A1 | |
| **Total** | **2** | |
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(RDDD) = 0.4 \times 0.3 \times 0.8 \times 0.8 = 0.0768,\ \frac{48}{625}$ | B1 | Correct probability for one clearly identified outcome evaluated; accept unsimplified. A correct unsimplified expression is not sufficient |
| $P(DRDD) = 0.6 \times 0.2 \times 0.3 \times 0.8 = 0.0288,\ \frac{18}{625}$ | M1 | Add 4 probability values, $0 < p < 1$, for appropriate identified scenarios. Accept unsimplified. Ways of identifying scenarios: stating the days; all unsimplified probability calculations exactly as stated; identifying correct branches on tree diagram and linking with values. No repeated scenarios. No incorrect scenarios |
| $P(DDRD) = 0.6 \times 0.8 \times 0.2 \times 0.3 = 0.0288,\ \frac{18}{625}$ | | |
| $P(DDDR) = 0.6 \times 0.8 \times 0.8 \times 0.2 = 0.0768,\ \frac{48}{625}$ | | |
| $0.2112,\ \frac{132}{625}$ | A1 | Accept 0.211. If 0/3 scored **SC B1** for $0.2112,\ \frac{132}{625}$ |
| **Total** | **3** | |
---2 A sports event is taking place for 4 days, beginning on Sunday. The probability that it will rain on Sunday is 0.4 . On any subsequent day, the probability that it will rain is 0.7 if it rained on the previous day and 0.2 if it did not rain on the previous day.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that it does not rain on any of the 4 days of the event.
\item Find the probability that the first day on which it rains during the event is Tuesday.
\item Find the probability that it rains on exactly one of the 4 days of the event.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q2 [6]}}